Question

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Given: DEFG is a parallelogram.

Parallelogram D E F G with diagonal D F. Angle G D F is labeled 1, angle F D E is labeled 2, angle E F D is labeled 3, and angle D F G is labeled 4.

Prove: DE¯¯¯¯¯≅GF¯¯¯¯¯ and DG¯¯¯¯¯¯≅EF¯¯¯¯¯ .

Answer 1

∠1 ≅ ∠3 and ∠2 ≅ ∠4 due to them being alternate interior angles.

ΔDGF ≅ ΔFED by the ASA Congruence Postulate.

How to prove a parallelogram is true?

From the given parallelogram, we can say that opposite sides of a parallelogram are equal and congruent and as such, we have;

∠1 ≅ ∠3 and ∠2 ≅ ∠4 due to them being alternate interior angles.

Secondly, we can say that DF is congruent to itself as DF ≅ DF due to the reflexive property of congruence.

Thirdly, we can say that ΔDGF ≅ ΔFED from ASA Congruence Postulate.

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